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Resistance is the opposition to the flow of current in a circuit. Understanding resistance and Ohm's law is fundamental to analysing any electrical circuit. This lesson covers the definition of resistance, the conditions under which Ohm's law applies, and the factors that affect the resistance of a conductor.
Spec mapping: This lesson maps to AQA 7408 specification section 3.5.1.3 — Resistance, covering the defining relationship R = V/I, the statement and conditions of Ohm's law, the distinction between Ohmic and non-Ohmic behaviour, and the qualitative dependence of resistance on temperature for metals, NTC thermistors and filament lamps. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- 3.5.1.4 — Resistivity: the R = V/I definition introduced here generalises in Lesson 5 to the material-property form R = ρL/A. The temperature-dependence arguments built here (lattice vibrations versus carrier concentration) carry forward unchanged.
- 3.2.2 — Quantum phenomena: the microscopic picture of free electrons scattering from lattice ions previews the quantum-mechanical band picture that distinguishes metals, semiconductors and insulators. Photoelectric questions and conduction questions share the same conceptual machinery of energy thresholds and carrier liberation.
- 3.6 — Thermal physics: the increase of metallic resistance with temperature is rooted in the equipartition argument from kinetic theory — higher T means greater amplitude of lattice vibration. The thermal-physics derivation of vibrational mean energy underwrites the qualitative claim made here.
Resistance is defined as the ratio of the potential difference across a component to the current flowing through it:
R=IV
where:
One ohm is the resistance of a component when a potential difference of one volt drives a current of one ampere through it (1 Ω = 1 V A⁻¹).
Key Point: The equation R = V/I is the definition of resistance. It applies to ALL components, whether or not they obey Ohm's law. Do not confuse this definition with Ohm's law itself.
Ohm's law states that the current through a conductor is directly proportional to the potential difference across it, provided the physical conditions (especially temperature) remain constant.
Mathematically: V ∝ I (at constant temperature)
This means V/I = constant, i.e., the resistance is constant.
A component that obeys Ohm's law is called an ohmic conductor. Its I-V graph is a straight line through the origin.
Ohm's law only holds when:
Most metallic conductors obey Ohm's law at constant temperature. Components like filament lamps, thermistors, and diodes do NOT obey Ohm's law because their resistance changes with current (due to temperature changes or their non-linear nature).
Common Misconception: Many students think V = IR is Ohm's law. It is not — V = IR is simply the definition of resistance rearranged. Ohm's law is the specific statement that V and I are proportional (constant R) under constant physical conditions.
The resistance of a uniform conductor depends on:
| Factor | Effect on resistance | Relationship |
|---|---|---|
| Length (L) | Longer wire → greater resistance | R ∝ L |
| Cross-sectional area (A) | Thicker wire → less resistance | R ∝ 1/A |
| Material (resistivity, ρ) | Different materials have different resistivities | R ∝ ρ |
| Temperature | Usually increases R for metals | Complex relationship |
These combine into the resistivity equation:
R=AρL
(This is covered in detail in Lesson 5 on Resistivity.)
A longer wire means charge carriers must travel further and undergo more collisions with the lattice ions. Each section of wire adds more resistance, so resistance is proportional to length.
A wider wire provides more "lanes" for charge carriers to flow through. Doubling the cross-sectional area is like having two identical resistors in parallel — each carries half the current, effectively halving the total resistance.
In metals, when temperature increases:
For metals, resistance increases approximately linearly with temperature over moderate ranges.
In semiconductors (and NTC thermistors), when temperature increases:
| Component | Temperature ↑ | Resistance |
|---|---|---|
| Metal wire | ↑ | ↑ increases |
| NTC thermistor | ↑ | ↓ decreases |
| Filament lamp | ↑ | ↑ increases (non-linear) |
Exam Tip: Always explain temperature effects in terms of the microscopic model — lattice vibrations and charge carrier behaviour. Simply stating "resistance goes up" without explaining why will not earn full marks.
A p.d. of 6.0 V is applied across a component. The current through it is 40 mA. Calculate the resistance.
Solution:
I = 40 mA = 0.040 A
R = V/I = 6.0 / 0.040 = 150 Ω
A 2.2 kΩ resistor is rated at 0.5 W maximum power. What is the maximum current it can safely carry?
Solution:
R = 2.2 kΩ = 2200 Ω
P = I²R, so I² = P/R
I² = 0.5 / 2200 = 2.27 × 10⁻⁴
I = √(2.27 × 10⁻⁴) = 0.0151 A = 15.1 mA
For the same 2.2 kΩ resistor rated at 0.5 W, what is the maximum voltage that can be applied?
Solution:
P = V²/R, so V² = PR
V² = 0.5 × 2200 = 1100
V = √1100 = 33.2 V
Check: P = V²/R = 1100/2200 = 0.5 W ✓
A metal wire has a resistance of 20.0 Ω at 20°C. When a large current flows through it, the wire heats up and its resistance increases to 23.5 Ω. The p.d. across the wire is maintained at 12 V. Calculate the current at (a) 20°C and (b) the higher temperature.
Solution:
(a) At 20°C: I = V/R = 12 / 20.0 = 0.600 A
(b) At higher temperature: I = V/R = 12 / 23.5 = 0.511 A
The current decreases because the resistance has increased due to the temperature rise.
Conductance (G) is the reciprocal of resistance:
G=R1=VI
The SI unit of conductance is the siemens (S), where 1 S = 1 Ω⁻¹ = 1 A V⁻¹.
A good conductor has high conductance and low resistance.
A component has a resistance of 250 Ω. Calculate its conductance.
Solution:
G = 1/R = 1/250 = 4.0 × 10⁻³ S = 4.0 mS
A student measures the I–V characteristic of an unknown component and obtains the following readings:
| V (V) | I (A) |
|---|---|
| 1.0 | 0.050 |
| 2.0 | 0.099 |
| 3.0 | 0.151 |
| 4.0 | 0.200 |
| 5.0 | 0.249 |
Calculate R = V/I at each voltage. Is the component Ohmic? Justify your answer.
Solution:
| V (V) | I (A) | R = V/I (Ω) |
|---|---|---|
| 1.0 | 0.050 | 20.0 |
| 2.0 | 0.099 | 20.2 |
| 3.0 | 0.151 | 19.9 |
| 4.0 | 0.200 | 20.0 |
| 5.0 | 0.249 | 20.1 |
The resistance is constant at ≈ 20 Ω within experimental scatter (the slight variations of ±0.2 Ω are well within the precision of typical lab meters). Since R is constant across a factor-of-five change in voltage, the component obeys Ohm's law over this range. To make this rigorous: plotting I against V would yield a straight line through the origin with gradient 1/R ≈ 0.050 A V⁻¹.
The corollary is important: a component is only Ohmic under the conditions tested. The same wire, driven hard enough to heat appreciably, would deviate from this linear behaviour at higher currents — Ohm's law is a limiting statement, not an unconditional truth.
A copper wire originally has length L = 1.0 m, cross-sectional area A = 1.0 × 10⁻⁶ m², and is measured to have a resistance of R₀ = 0.017 Ω at 20 °C. The wire is now replaced with a new sample of the same material but with: (i) length doubled to 2.0 m, and (ii) cross-sectional area halved to 0.5 × 10⁻⁶ m². The new wire is then heated to 70 °C. The temperature coefficient of resistivity for copper is α ≈ 4.0 × 10⁻³ K⁻¹. (a) Predict the new resistance at 20 °C from the geometry alone. (b) Predict the resistance at 70 °C. (c) Comment on which factor — geometry or temperature — dominates the change.
Solution — Part (a): Geometric scaling at constant temperature.
The defining relation is R = ρL / A. Holding ρ constant (same material, same temperature), the ratio of resistances is:
R_new / R₀ = (L_new / L₀) × (A₀ / A_new)
Substituting: R_new / R₀ = (2.0 / 1.0) × (1.0 × 10⁻⁶ / 0.5 × 10⁻⁶) = 2 × 2 = 4
So R_new (at 20 °C) = 4 × 0.017 = 0.068 Ω.
Conceptual check: Doubling the length doubles the path the electrons must traverse, doubling the resistance. Halving the area concentrates the same current through half the cross-section, doubling the resistance again. The two effects multiply, giving a factor of 4. Many students incorrectly add the two factors (2 + 2 = 4 happens to give the right answer here, but only by numerical coincidence). The correct reasoning is multiplicative.
Part (b): Adding the temperature correction.
The temperature coefficient of resistivity α describes how the resistivity (and hence resistance) of a metal varies linearly with temperature over a moderate range:
R(T) = R(T₀) × [1 + α(T − T₀)]
With T − T₀ = 70 − 20 = 50 K and α = 4.0 × 10⁻³ K⁻¹:
Correction factor = 1 + (4.0 × 10⁻³ × 50) = 1 + 0.20 = 1.20
R(70 °C) = 0.068 × 1.20 = 0.0816 Ω ≈ 0.082 Ω (to 2 s.f.)
Part (c): Which factor dominates?
Compare the multiplicative effects:
The geometric change overwhelmingly dominates. This is generally true for everyday electrical engineering: changing the physical dimensions of a conductor produces much larger fractional changes in resistance than typical operating-temperature swings. The exception is in components such as filament lamps, where the conductor temperature jumps from ~20 °C to ~2500 °C — across that range the temperature term dominates instead, and the cold-to-hot resistance ratio can exceed 10.
Note: A-Level papers occasionally test the inverse problem: given measured resistances at two temperatures, deduce α. The same equation R(T) = R(T₀)[1 + α(T − T₀)] handles both directions. Worth memorising that for pure copper at room temperature α ≈ 4 × 10⁻³ K⁻¹, while for the alloys typically used in standard resistors (constantan, manganin) α is roughly 100× smaller — that is exactly why precision resistors use alloys, not pure copper.
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